Construction of Components of Second Bimedial

From ProofWiki
Jump to navigation Jump to search


In the words of Euclid:

To find medial straight lines commensurable in square only which contain a medial rectangle.

(The Elements: Book $\text{X}$: Proposition $28$)


Let $\rho$, $\rho \sqrt k$ and $\rho \sqrt \lambda$ be rational straight lines which are commensurable in square only.

Take the mean proportional $\rho \sqrt [4] k$ of $\rho$ and $\rho \sqrt k$, which is medial.

Let $x$ be such that:

$\rho \sqrt k : \rho \sqrt \lambda = \rho \sqrt [4] k : x$

which gives:

$x = \dfrac {\rho \sqrt \lambda} {\sqrt [4] k}$

We have that:

$\rho \sqrt k \frown \!\! - \rho \sqrt \lambda$

where $\frown \!\! -$ denotes commensurability in square only.


$\rho \sqrt [4] k \frown \!\! - x$

From Straight Line Commensurable with Medial Straight Line is Medial it follows that $\rho k^{3/4}$ is also medial.


\(\ds \rho \sqrt \lambda : x\) \(=\) \(\ds \rho \sqrt k : \rho \sqrt [4] k\)
\(\ds \) \(=\) \(\ds \rho \sqrt [4] k : \rho\)
\(\ds \leadsto \ \ \) \(\ds x \cdot \rho \sqrt [4] k\) \(=\) \(\ds \rho^2 \sqrt \lambda\) which is a medial area


Also see

Historical Note

This proof is Proposition $28$ of Book $\text{X}$ of Euclid's The Elements.